Question

A ball A is projected from origin with an initial
velocity V = 700 cm/s, in a direction 37 degree above
the horizontal as shown in fig. Another ball B
300 cm from origin on a line 37 degree above the
horizontal is released from rest at the instant A
starts. then how far will B have fallen when it is
hit by A-
(A) 80 cm
(B) 60 cm
(C) 90 cm
(D) 70 cm
sorry for the inconvenience. this is the full question.​

Answer 1

Answer:

A. 80

Explanation:

Please refer to attachment .

Answer 2

Given:

Initial velocity of A= 700 cm/s

The angle of projection of A and B = 37°

Height of projection of B (AB) = 300cm

To find:

Distance covered by B when it was hit by A

Solution:

x = AB cos 37° = AB X 4/5     (using trignometry)

= 300 X 4/5 = 240cm

time = distance / speed

or t = 2.4m / 7 ms⁻¹

Applying equation of motion,

y = ut +1/2 gt²

Since initial velocity of B = 0m/s,

y = 1/2 gt²

Substituting,

y = 10 X 2.4 X 2.4 / 2 X 7 X 7

= 60 cm

Hence, the distance covered is 60cm.